How much ice to float a 80 Kg person

[iceman_ch]

1. The problem statement, all variables and given/known data

Find the volume of the smallest block of ice which when floating on salt water will carry a person of mass 80 Kg

Densities:

Salt Water = 1025
Ice = 917

2. Relevant equations

F_b=P_f * V_f * G

3. The attempt at a solution

I figured F_b would have to be equal to the force of the person down which is equal to 784 N. Then plugging that into the equation and solving for V_f I get .078. I used the Salt Water for P_f. Is there somewhere that I need to plug in Ice?

[Hootenanny]

Don't forget the weight of the ice...:wink:

[iceman_ch]

ok so I tried to add the weight of ice in here is what I get.

like before I have the equation.

W_p * G = 784

This is the force of the pearson. Next I have

784 = Volume of the ice * Density of water * Gravity

Then to add in the weight of ice I do this.

Volume of the ice * Density of Ice * Gravity + 784 = Volume of ice * density of water * gravity

This makes an equation has everything cancel out.
What am I doing wrong?

[denverdoc]

ok so I tried to add the weight of ice in here is what I get.

like before I have the equation.

W_p * G = 784

This is the force of the pearson. Next I have

784 = Volume of the ice * Density of water * Gravity

Then to add in the weight of ice I do this.

Volume of the ice * Density of Ice * Gravity + 784 = Volume of ice * density of water * gravity

This makes an equation has everything cancel out.
What am I doing wrong?

(vol ice*(density ice)+m)g=volume ice(density SALT water)*g Now the question is ice salty? I think not. On the left hand side we have the total weight of the ice plus man, on the right the weight of water displaced.